Question

this is linear algebra & differential equation about Homogenous Equation question...
convert to separable eqs: v=y/x

1.) xyy' = 2y^2-x^2

Answer 1

change y' to dy/dx
\[\implies (xy)\frac{dy}{dx} = 2y^2 - x^2\]
cross multiply..
\[\implies (xy)dy = (2y^2 - x^2)dx\]
put everything on one side...
\[\implies (2y^2- x^2)dx - (xy)dy = 0\]
let y = vx

Answer 2

so dy = vdx + xdv

Answer 3

\[\implies (2v^2 x^2 - x^2)dx - (vx^2)(vdx + xdv) = 0\]
multiply the thingies...
\[\implies 2v^2 x^2 dx - x^2 dx - v^2 x^2 dx - vx^3 dv = 0\]
are you still following?

Answer 4

i think @adunb8 went offline o.O

Answer 5

yea im on sorry

Answer 6

so you get what im doing so far?

Answer 7

yes! omg! but teacher wants me to determine if its separable or not then use the v=y/x so how do determine it is separable eqn?

Answer 8

all homogeneous are separable in the later stage...which im going to do now

Answer 9

okay but how do i determine that on the test because on the test he will only say use whatever your learned so far and he wont specify which to use

Answer 10

first combine like terms...
\[\implies v^2 x^2 dx -x^2 dx - vx^3 dv\]
factor out
\[\implies x^2dx(v^2 - 1) - vx^3dv\]
do you see the separable thingy now?

Answer 11

i can tell you how to know if it's homogeneous...

Answer 12

okay=)

Answer 13

you look at the degree...when the degree are all the same...it's homogeneous

for example

(xy) <--degree 2

(2y^2) <--degree 2

(x^2) <--degree 2

therefore, it's homogeneous

Answer 14

why is (xy) degree of 2?

Answer 15

x^1 y^1 <--1 + 1 = 2

Answer 16

oh i see sorry i dont really see things right away...

Answer 17

note: ALL terms have to be SAME degree for it to be homo

Answer 18

i see!!

Answer 19

for exact de \[\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\]

Answer 20

for linear DE \[\frac{dy}{dx} + P(x) y = Q(x)\]

Answer 21

what about for separable?

Answer 22

i am not aware of any mark for it. i think it's just trial and error

Answer 23

oh okay... i guess that really has to be trial and error

Answer 24

but usually it's in the form \[f(x)dx + g(y)dy = 0\]

Answer 25

that means dx only has function of x and dy only has function of y

Answer 26

however, there are times, when you can manipulate an equation into becoming separable

Answer 27

okay so after i plug into the v= y/x formula then what do i do in order to solve the final solution?

Answer 28

you dont do v = y/x yet

like i said your FIRST substitution is y = vx

Answer 29

you do v = y/x in the back subsitution... AFTER you get the solution

Answer 30

okay so after y=vx i plug back into the original solution?

Answer 31

are we talking about the first step?

Answer 32

im talking about here you said above couple of replies earlier.
first combine like terms...
⟹v2x2dx−x2dx−vx3dv

factor out
⟹x2dx(v2−1)−vx3dv

do you see the separable thingy now?

Answer 33

oh.. now you do variable separable

Answer 34

you solve for the solution

Answer 35

i think i need to solve for y but i see that you changed all of them to y =vx

Answer 36

you dont solve for anything yet....

Answer 37

you just solve for the GENERAL SOLUTION

Answer 38

how do i do that from here? use by doing integral steps?

Answer 39

like i said...variable separable

Answer 40

i dont really understand what you mean by variable separable..... sorry =(

Answer 41

\[x^2 dx(v^2 - 1) - vx^3 dv\]
\[\implies \frac{x^2dx}{x^3} - \frac{vdv}{v^2 - 1}\]
that's what variable separation means

Answer 42

oh isee okay !

Answer 43

so solve the general solution now

Answer 44

by doing the integral right? and using u subsitution?

Answer 45

yes

Answer 46

thank you so so much you are my life saver every time! !! =)

Answer 47

welcome

Answer 48

after you integrate..change v into y/x

Answer 49

then that would be the final answer

Answer 50

right back to original!

Answer 51

yep